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Thread: Help with Proof of Sets

  1. #1
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    Help with Proof of Sets

    Prove $\displaystyle (A \cup B) - (A \cap B) = (A-B) \cup (B-A)$

    $\displaystyle (A \cup B) - (A \cap B) \subseteq (A-B) \cup (B-A)$
    Let $\displaystyle x \in (A \cup B) - (A \cap B)$. Then $\displaystyle x \in A \cup B$ and $\displaystyle x \notin A \cap B$. Then $\displaystyle x \in A$ or $\displaystyle x \in B$ but $\displaystyle x \notin A$ and $\displaystyle x \notin B$. Since $\displaystyle x \in A$ and $\displaystyle x \notin B$ then $\displaystyle x \in (A-B)$. Also since $\displaystyle x \in B$ and $\displaystyle x \notin A$ then, $\displaystyle x \in (B-A)$. Now $\displaystyle x \in (A-B)\cup(B-A)$.

    Therefore, $\displaystyle (A \cup B) - (A \cap B) \subseteq (A-B) \cup (B-A)$

    Do I have the general idea right? What can I fix? I'm sure I did something wrong. Yes, I also realized I skipped the other part of the proof.
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  2. #2
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    Hello, JSB1917!

    I used a two-column proof . . .


    $\displaystyle \text{Prove: }\;(A \cup B) - (A \cap B) = (A-B) \cup (B-A)$

    $\displaystyle \begin{array}{ccccccc}
    1. & (A \cup B) - (A \cap B) && 1.& \text{ Given} \\ \\
    2. & (A \cup B) \cap (\overline{A \cap B}) && 2. & \text{d{e}f. Subtraction} \\ \\
    3. & (A \cup B) \cap (\overline A \cup \overline B) && 3. & \text{DeMorgan's Law} \\ \\
    4. & (A \cap \overline A) \cup (A \cap \overline B) \cup (B \cap \overline A) \cup (B \cap \overline B) && 4. & \text{Distributive Prop.} \\ \\
    5. & \emptyset \;\;\cup (A \cap \overline B) \cup (B \cap \overline A) \cup \;\;\emptyset && 5. & S \cap S \:=\:\emptyset \\ \\
    6. & (A \cap \overline B) \cup (B \cap \overline A) && 6. & S \cup \emptyset \:=\:S \\ \\
    7. & (a - B) \cup (B - A) && 7. & \text{d{e}f. Subtraction}
    \end{array}$
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